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26 May 2017, 04:43
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If \(x = -\frac{1}{\sqrt{0.91}}, y = -\frac{1}{0.91^{2}},\) and \(z = -\frac{1}{0.91}\), which of the following is true?
A. \(y < z < x\)
B. \(y < x < z\)
C. \(x < z < y\)
D. \(z < x < y\)
E. \(z < y < x\)
A. \(y < z < x\)
B. \(y < x < z\)
C. \(x < z < y\)
D. \(z < x < y\)
E. \(z < y < x\)
26 May 2017, 04:46
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Official Solution:
If \(x = -\frac{1}{\sqrt{0.91}}, y = -\frac{1}{0.91^{2}},\) and \(z = -\frac{1}{0.91}\), which of the following is true?
A. \(y < z < x\)
B. \(y < x < z\)
C. \(x < z < y\)
D. \(z < x < y\)
E. \(z < y < x\)
In general, if \(a > b > 0\), you get \(\frac{1}{a} < \frac{1}{b}\), and if you multiply the negatives on both sides, the sign of inequality changes again, and you get \(-\frac{1}{a} > -\frac{1}{b}\).
In other words, if you compare \(a > b > 0\) and \(-\frac{1}{a} > -\frac{1}{b}\), you can clearly see that the sign of both inequalities ">" are the same. Thus, if you want to find which of \(-\frac{1}{a}\) or \(-\frac{1}{b}\) is greater, you need to find which of a or b is greater. Therefore, the order of size of \(x = -\frac{1}{\sqrt{0.91}}, y = -\frac{1}{0.91^{2}},\) and \(z = -\frac{1}{0.91}\) is the same order of size as \(\sqrt{0.91}, 0.91^{2}\), and 0.91. However, if \(0 < x < 1\), you get \(x^{2} < x < \sqrt{x}\). Using substitution, you get \(0.91^{2} < 0.91 < \sqrt{0.91}\), and this is the same as the size order of \(y, z, x,\) so you get \(y < z < x\). The answer is A.
Answer: A
If \(x = -\frac{1}{\sqrt{0.91}}, y = -\frac{1}{0.91^{2}},\) and \(z = -\frac{1}{0.91}\), which of the following is true?
A. \(y < z < x\)
B. \(y < x < z\)
C. \(x < z < y\)
D. \(z < x < y\)
E. \(z < y < x\)
In general, if \(a > b > 0\), you get \(\frac{1}{a} < \frac{1}{b}\), and if you multiply the negatives on both sides, the sign of inequality changes again, and you get \(-\frac{1}{a} > -\frac{1}{b}\).
In other words, if you compare \(a > b > 0\) and \(-\frac{1}{a} > -\frac{1}{b}\), you can clearly see that the sign of both inequalities ">" are the same. Thus, if you want to find which of \(-\frac{1}{a}\) or \(-\frac{1}{b}\) is greater, you need to find which of a or b is greater. Therefore, the order of size of \(x = -\frac{1}{\sqrt{0.91}}, y = -\frac{1}{0.91^{2}},\) and \(z = -\frac{1}{0.91}\) is the same order of size as \(\sqrt{0.91}, 0.91^{2}\), and 0.91. However, if \(0 < x < 1\), you get \(x^{2} < x < \sqrt{x}\). Using substitution, you get \(0.91^{2} < 0.91 < \sqrt{0.91}\), and this is the same as the size order of \(y, z, x,\) so you get \(y < z < x\). The answer is A.
Answer: A
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